Inverse Trigonometric Functions

15 min read

A surveyor stands a known distance from a tower and measures how far up the top sits relative to eye level. The ratio of height to distance is fixed by the geometry, but the surveyor does not want the ratio. The surveyor wants the angle of elevation. Sine, cosine, and tangent each turn an angle into a ratio. The question “I have the ratio, what was the angle?” runs that machine backward. That backward direction is the inverse trigonometric function, and it is the tool behind a robot arm computing a joint angle from a reach, a camera computing an aim from a height, and the entire calculus integral $\int \frac{dx}{1+x^2}$.

The one obstacle, and it is the heart of the topic: sine, cosine, and tangent each repeat, so a single ratio is produced by infinitely many angles. A function is allowed only one output per input. To get a genuine inverse, each trig function is restricted to one piece on which it never repeats a value, and only that piece is reversed. The chosen piece is the range of the inverse, and the range is the rule that decides which one angle the inverse is allowed to report.

By the end of this page you can:

• State, from memory, the domain and range of $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$.
• Evaluate a direct inverse value such as $\cos^{-1}\!\left(-\frac{\sqrt{2}}{2}\right)$ by finding the one angle inside the range.
• Evaluate a composition such as $\sin^{-1}\!\left(\sin\frac{5\pi}{6}\right)$ where the inner angle falls outside the range.
• Build a reference triangle to simplify a compound expression such as $\cos\!\left(\sin^{-1} x\right)$.
• Tell $\sin^{-1} x$ (an angle) apart from $\frac{1}{\sin x}$ (a ratio).

The plain floor: an inverse trig function takes a ratio and returns one angle, and that angle must lie in a fixed narrow range. For inverse sine the range is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, for inverse cosine it is $[0, \pi]$, and for inverse tangent it is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. If a remembered answer falls outside the range, it is the wrong angle, and the right angle is the one with the same ratio that does lie in the range. Memorizing those three ranges and applying that one test covers most of the graded work.

Quick Reference

Before You Start

Check each box you can do from memory. A box you cannot check yet points to a quick refresher, not a grade.

• I can read a special value off the unit circle, such as $\sin\frac{\pi}{6}$ or $\cos\frac{3\pi}{4}$. (The Unit Circle: Sine and Cosine)
• I can explain why the full sine function fails the horizontal line test.
• I can find the third side of a right triangle from two sides using the Pythagorean theorem. (Right Triangle Trigonometry)

A 60-second self-check. Pick an answer, then reveal the reasoning.

Try This First

Before any formula, work this with what you already know about the unit circle.

You know that $\sin\!\left(\frac{\pi}{6}\right) = \frac{1}{2}$. Turn that around: which angle has a sine of $\frac{1}{2}$?

The Idea Behind Inverse Trig Functions

Sine takes an angle and returns a ratio. The inverse sine reverses that process: it takes a ratio and returns an angle. As the input ratio grows from $-1$ toward $1$, the output angle grows from the bottom of a chosen range toward the top of it.

There is a complication, and it is the heart of the topic. Sine, cosine, and tangent repeat. Each output value is produced by infinitely many inputs. A reversal that returns “all the angles” is not a function. To get a genuine inverse, each trig function is restricted to a piece of its domain on which it never repeats a value, and only that piece is reversed. The restricted piece is chosen so the reversal still covers every possible ratio. That single chosen piece is what makes $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$ each return one specific angle.

So the inverse trig functions are functions with a fixed, narrow range. The range is not a detail to memorize and forget. It is the rule that decides which of the infinitely many candidate angles the inverse is allowed to report.

Prerequisite Hub

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Builds on:

Cross-course prerequisite: Inverse Functions (math141-inverse-functions, MATH 141). The general rule that a function must be one-to-one before it can be inverted, and that inverse graphs reflect across $y = x$, comes from this MATH 141 skill.

Unlocks (what these inverses feed into next):

In calculus the inverse trig functions reappear as antiderivatives, so the ranges fixed here become the ranges of those antiderivatives. That is tracked in a later course.

Quick Reference

Read the range as a rule: the output of $\sin^{-1}$ is always an angle in the right two quadrants ($-90^\circ$ to $90^\circ$); the output of $\cos^{-1}$ is always an angle in the top two quadrants ($0^\circ$ to $180^\circ$); the output of $\tan^{-1}$ is strictly between $-90^\circ$ and $90^\circ$.

Notation warning: here $\sin^{-1} x$ means the inverse sine of $x$. It does not mean $\frac{1}{\sin x}$. The reciprocal of sine is written $\csc x$, never $\sin^{-1} x$.

The three ranges are the most-confused part of the topic. A quick check before the worked examples.

Official Definitions

Each inverse below restricts its trig function to a one-to-one piece, then reverses that piece. The restricted ranges match Stewart 7e, Section 5.5, with the OpenStax alternate covering the same definitions in Section 6.3.

Inverse sine function. $y = \sin^{-1} x$ (also written $\arcsin x$) has domain $[-1, 1]$ and range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. On the restricted domain $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, if $\sin y = x$ then $\sin^{-1} x = y$.

Stewart 7e, Section 5.5 Inverse Trigonometric Functions and Their Graphs; OpenStax Precalculus 2e, Section 6.3.

Inverse cosine function. $y = \cos^{-1} x$ (also written $\arccos x$) has domain $[-1, 1]$ and range $[0, \pi]$. On the restricted domain $[0, \pi]$, if $\cos y = x$ then $\cos^{-1} x = y$.

Stewart 7e, Section 5.5 Inverse Trigonometric Functions and Their Graphs; OpenStax Precalculus 2e, Section 6.3.

Inverse tangent function. $y = \tan^{-1} x$ (also written $\arctan x$) has domain $(-\infty, \infty)$ and range $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. On the restricted domain $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, if $\tan y = x$ then $\tan^{-1} x = y$.

Stewart 7e, Section 5.5 Inverse Trigonometric Functions and Their Graphs; OpenStax Precalculus 2e, Section 6.3.

Two representations of the same idea

The definition above is the algebraic representation. Here is the graphical one for inverse sine, the reflection of the restricted sine graph across the line $y = x$.

   y
   π/2 ┤                          ●  (1, π/2)
       │                      ●
       │                  ●
     0 ┼──────────────●────────────── x
       │          ●
       │      ●
  -π/2 ┤  ●  (-1, -π/2)
       └──┬──────┬──────┬──────┬──
         -1            0            1

The domain runs from $-1$ to $1$ along the horizontal axis (every legal input ratio). The range runs from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ along the vertical axis (every angle the function is allowed to return). Reading left to right, as the ratio increases the output angle increases without ever repeating. That is exactly what makes the reversal a function.

Translate between the two: the definition says “if $\sin y = x$ then $\sin^{-1} x = y$ on the restricted range.” Find the point $\left(\frac{1}{2}, \frac{\pi}{6}\right)$ on the graph and confirm it says the same thing: input $\frac{1}{2}$, output $\frac{\pi}{6}$.

Now check the two ranges that students most often swap.

Key Theorem: Composition Properties

Inverse (composition) properties of trig functions. $\sin\!\left(\sin^{-1} x\right) = x$ for $-1 \le x \le 1$; $\cos\!\left(\cos^{-1} x\right) = x$ for $-1 \le x \le 1$; $\tan\!\left(\tan^{-1} x\right) = x$ for all real $x$. In the other order, $\sin^{-1}\!\left(\sin x\right) = x$ only for $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$; $\cos^{-1}\!\left(\cos x\right) = x$ only for $0 \le x \le \pi$; $\tan^{-1}\!\left(\tan x\right) = x$ only for $-\frac{\pi}{2} < x < \frac{\pi}{2}$.

Stewart 7e, Section 5.5 Inverse Trigonometric Functions and Their Graphs; OpenStax Precalculus 2e, Section 6.3.

Why the order matters

One direction always cancels; the other has a condition attached. The reason is the restricted range.

When you compute $\sin\!\left(\sin^{-1} x\right)$, the inner step first lands on an angle inside the allowed range, and then sine sends that angle back to the ratio $x$. Nothing can go wrong, so the cancellation holds for every legal $x$.

When you compute $\sin^{-1}\!\left(\sin x\right)$, the inner step may hand the outer function an angle that lies outside the allowed range. The outer $\sin^{-1}$ is required to report an answer inside $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so it returns the angle in that range with the same sine, which need not be the $x$ you started with. The equality holds only when $x$ was already in the range.

Worked Examples

Example 1: A direct inverse value (predict, then check)

Evaluate $\cos^{-1}\!\left(-\frac{\sqrt{2}}{2}\right)$.

Predict first. The input is negative, and the range of $\cos^{-1}$ is $[0, \pi]$. Cosine is negative in the second quadrant, so the answer should be an angle between $\frac{\pi}{2}$ and $\pi$. Predict an answer in that band before computing.

Compute. You are looking for the angle $y$ in $[0, \pi]$ with $\cos y = -\frac{\sqrt{2}}{2}$. The reference angle with cosine $\frac{\sqrt{2}}{2}$ is $\frac{\pi}{4}$. The second-quadrant angle with that reference angle is

$$y = \pi - \frac{\pi}{4} = \frac{3\pi}{4}.$$

Check the prediction. $\frac{3\pi}{4}$ is between $\frac{\pi}{2}$ and $\pi$, matching the prediction. Confirm the value: $\cos\!\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$. Correct.

$$\cos^{-1}\!\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}.$$

Example 2: The composition that does not simply cancel

Evaluate $\sin^{-1}\!\left(\sin\!\left(\frac{5\pi}{6}\right)\right)$.

Predict first. It is tempting to answer $\frac{5\pi}{6}$. Before accepting that, check whether $\frac{5\pi}{6}$ is inside the range of $\sin^{-1}$, which is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Since $\frac{5\pi}{6} > \frac{\pi}{2}$, it is not in the range, so the answer cannot be $\frac{5\pi}{6}$. Predict instead an angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ that has the same sine.

Compute. First find the inner value: $\sin\!\left(\frac{5\pi}{6}\right) = \frac{1}{2}$. Now evaluate $\sin^{-1}\!\left(\frac{1}{2}\right)$, the angle in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ whose sine is $\frac{1}{2}$:

$$\sin^{-1}\!\left(\frac{1}{2}\right) = \frac{\pi}{6}.$$

Check the prediction. $\frac{\pi}{6}$ is inside the allowed range, and it has the same sine as $\frac{5\pi}{6}$. The composition did not return the original angle, exactly as the theorem warned.

$$\sin^{-1}\!\left(\sin\!\left(\frac{5\pi}{6}\right)\right) = \frac{\pi}{6}.$$

This is the moment the inner-versus-outer order matters. Check that you can spot when a composition cancels and when it does not.

Example 3: A composition that builds a triangle

Evaluate $\cos\!\left(\sin^{-1}\!\left(\frac{3}{5}\right)\right)$.

Predict first. The inner part, $\sin^{-1}\!\left(\frac{3}{5}\right)$, is a positive angle in the first quadrant (its sine is positive and its range puts it in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$). Cosine of a first-quadrant angle is positive, so predict a positive answer less than $1$.

Compute. Let $\theta = \sin^{-1}\!\left(\frac{3}{5}\right)$, so $\sin\theta = \frac{3}{5}$ with $\theta$ in the first quadrant. Build a reference right triangle: opposite side $3$, hypotenuse $5$. The adjacent side is

$$\sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4.$$

Then

$$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5}.$$

Check the prediction. $\frac{4}{5}$ is positive and less than $1$, matching the prediction. As a sanity check, $\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 = \frac{9}{25} + \frac{16}{25} = 1$, so $\sin^2\theta + \cos^2\theta = 1$ holds.

$$\cos\!\left(\sin^{-1}\!\left(\frac{3}{5}\right)\right) = \frac{4}{5}.$$

Example 4: Two valid paths, one answer

Evaluate $\tan^{-1}\!\left(\tan\!\left(\frac{3\pi}{4}\right)\right)$.

One way to see this. Compute the inner value: $\tan\!\left(\frac{3\pi}{4}\right) = -1$. Then $\tan^{-1}(-1)$ is the angle in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with tangent $-1$, which is $-\frac{\pi}{4}$.

Another way to see this. Use the theorem directly. The cancellation $\tan^{-1}\!\left(\tan x\right) = x$ holds only for $x$ in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Here $x = \frac{3\pi}{4}$ is outside that interval, so subtract a half-period $\pi$ (tangent repeats every $\pi$) to land in range: $\frac{3\pi}{4} - \pi = -\frac{\pi}{4}$.

Both paths give

$$\tan^{-1}\!\left(\tan\!\left(\frac{3\pi}{4}\right)\right) = -\frac{\pi}{4}.$$

The first path is mechanical and safe. The second is faster once the range condition is second nature. Either is correct.

Common Misconceptions

Practice Problems

Go Deeper (optional)

None of this is needed to evaluate inverse trig values or to pass the section. It is here for readers who want to see where the inverse functions lead.

The inverse tangent is how machines recover an angle from a measurement. A robot arm knows the horizontal reach $x$ and the vertical reach $y$ of its hand and needs the joint angle. That angle is $\tan^{-1}\!\left(\frac{y}{x}\right)$. The reason $\tan^{-1}$ accepts every real number, unlike $\sin^{-1}$ and $\cos^{-1}$, is that a height-to-distance ratio can be any size, while a sine or cosine is trapped in $[-1, 1]$. This is the same computation a camera uses to aim at a tracked object and a navigation system uses to find a heading from north and east components.

The ranges become the ranges of antiderivatives in calculus. Several standard integrals evaluate to inverse trig functions, for example $\int \frac{dx}{\sqrt{1 - x^2}} = \sin^{-1} x + C$ and $\int \frac{dx}{1 + x^2} = \tan^{-1} x + C$. The narrow ranges fixed on this page are what make those antiderivatives single-valued functions rather than ambiguous expressions. A later course develops this, but the range rule learned here is exactly the rule that carries over.

Check Yourself

Close the notes and answer each from memory, then reveal it. Pulling an idea back from memory is one of the strongest ways to make it stick.

Mastery Checklist

Novice (Level 1-2):

• I can state the domain and range of $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$ from memory.
• I can explain why each trig function must be restricted before it can be inverted.
• I can evaluate a direct inverse value such as $\sin^{-1}\!\left(-\frac{1}{2}\right)$.

Competent (Level 3-4):

• I can evaluate compositions where the inner angle falls outside the range and report the in-range answer.
• I can build a reference triangle to evaluate a composition such as $\cos\!\left(\sin^{-1} x\right)$ at a number.
• I can distinguish $\sin^{-1} x$ from $\frac{1}{\sin x}$, and the range of $\sin^{-1}$ from the range of $\cos^{-1}$.

Proficient (Level 5):

• I can derive a general algebraic identity for a composition and justify its sign from the range.
• I can explain why one order of composition cancels for all valid inputs while the other has a condition.

Mental Model

Think of each inverse trig function as a strict tour guide. Many angles share the same sine, cosine, or tangent, but the guide is only allowed to take you to one specific neighborhood: the right two quadrants for inverse sine, the top two for inverse cosine, the open right two for inverse tangent. Whatever ratio you bring, the guide returns the one angle in its assigned neighborhood. That is why a composition such as $\sin^{-1}\!\left(\sin x\right)$ can hand you back a different angle than the one you started with: the guide can only return an angle from its own neighborhood.

Connections

Looking back:

• The Unit Circle: Sine and Cosine supplies the special-angle values used throughout these evaluations.
• Right Triangle Trigonometry supplies the reference-triangle method for compositions.
• Inverse Functions (MATH 141) supplies the general rule that a function must be one-to-one before it can be inverted.

Looking ahead:

• Solving trigonometric equations uses inverse trig functions to recover angles, then adds full periods to list all solutions.
• Calculus integrates several standard expressions to inverse trig functions, so the ranges here become the ranges of antiderivatives.

Real-world connection:

• A camera tracking an object computes the aiming angle as the inverse tangent of the height-to-distance ratio, which is why $\tan^{-1}$ accepts every real input.

Resources

All open resources are free and openly licensed. The definitions and the composition theorem on this page follow Stewart 7e, Section 5.5, and OpenStax Precalculus 2e, Section 6.3, and the restricted ranges match both sources.

Last updated: 2026-06-24